In a laboratory experiment, 3.45 g of Sr(NO3)2 is dissolved in enough water to make exactly 750 mL of solution. A 100.0 mL sample of this solution is titrated with a 0.0460 M solution of Na2CrO4-

Question

What volume of Na2CrO4 solution is required to precipitate all of the SrCrO4

Answer 1

Sr(NO3)2(aq)+Na2CrO4(aq)-->SrCrO4(s)+2NaNO3(aq)

Find the molarity of Sr(NO3)2:

3.45g/211.628g/mol=0.016302mol

M=0.016302mol/0.750L=0.021736 M

100 mL Sr(NO3)2 to mL of Na2CrO4

mL to mol with molarity:

100mL Sr(NO3)2 (0.021736mol/1000mL)=0.0021736 mol Sr(NO3)2

mol to mol with coefficients:

0.0021736 mol Sr(NO3)2 (1 mol Na2CrO4 / 1 mol Sr(NO3)2 ) = 0.0021736 mol Na2CrO4

mol to mL with molarity:

0.0021736 mol Na2CrO4 (1000mL/0.0460 mol)=47.252=47.3 mL